An experiment is rolling two fair dice. a) What is the probability that the sum is 8 ? Round answer to 3 decimal places. \( \square \) b) What is the probability that the sum is 8 given that the first die is a 3 ? Round answer to 3 decimal places. \( \square \) c) What is the probability that the first die is a 3 given that the sum is 8 ? Round answer to 3 decimal places. \( \square \) d) Are the events the sum is 8 and the first die is a 3 independent events? Why or why not? Round answer to 3 decimal places. Select an answer e) What is the probability that the sum is 7 ? Round answer to 3 decimal places. \( \square \) f) What is the probability that the sum is 7 given that the first die is a 2? Round answer to 3 decimal places. \( \square \) g) What is the probability that the first die is a 2 given that the sum is \( 7 ?^{\circ} \) Round answer to 3 decimal places. \( \square \) h) Are the events the sum is 7 and the first die is a 2 independent events?

a) There are 36 equally likely outcomes when rolling two dice. The outcomes with a sum of 8 are \[ (2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2) \] There are 5 such outcomes. Thus, the probability is \[ P(\text{sum} = 8) = \frac{5}{36} \approx 0.139. \] b) Given that the first die is a 3, the sample space reduces to the outcomes \[ (3,1),\ (3,2),\ (3,3),\ (3,4),\ (3,5),\ (3,6) \] There are 6 outcomes. The only outcome where the sum is 8 is \((3,5)\) because \(3+5=8\). Thus, \[ P(\text{sum}=8 \mid \text{first die}=3) = \frac{1}{6} \approx 0.167. \] c) When the sum is 8, the possible outcomes are \[ (2,6),\ (3,5),\ (4,4),\ (5,3),\ (6,2) \] There are 5 outcomes. Only one of these, \((3,5)\), has the first die equal to 3. Hence, \[ P(\text{first die}=3 \mid \text{sum}=8) = \frac{1}{5} = 0.200. \] d) Two events \(A\) (the sum is 8) and \(B\) (the first die is 3) are independent if \[ P(A \text{ and } B) = P(A) \times P(B). \] We have \[ P(A) = \frac{5}{36} \quad \text{and} \quad P(B) = \frac{6}{36} = \frac{1}{6}. \] Thus, \[ P(A) \times P(B) = \frac{5}{36} \times \frac{1}{6} = \frac{5}{216} \approx 0.023. \] But previously, we found \[ P(A \text{ and } B) = P((3,5)) = \frac{1}{36} \approx 0.028. \] Since \(\frac{1}{36} \neq \frac{5}{216}\), the events are not independent. e) The outcomes with a sum of 7 are \[ (1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1) \] There are 6 outcomes. Hence, \[ P(\text{sum}=7) = \frac{6}{36} = \frac{1}{6} \approx 0.167. \] f) Given that the first die is 2, the possible outcomes are \[ (2,1),\ (2,2),\ (2,3),\ (2,4),\ (2,5),\ (2,6) \] For the sum to be 7, we need \(2+{\text{second die}}=7\) so that the second die must be 5. There is 1 outcome. Therefore, \[ P(\text{sum}=7 \mid \text{first die}=2) = \frac{1}{6} \approx 0.167. \] g) When the sum is 7, the outcomes are \[ (1,6),\ (2,5),\ (3,4),\ (4,3),\ (5,2),\ (6,1) \] Out of these 6 outcomes, only one has the first die equal to 2: \((2,5)\). Thus, \[ P(\text{first die}=2 \mid \text{sum}=7) = \frac{1}{6} \approx 0.167. \] h) To check if the events "the sum is 7" and "the first die is 2" are independent, we verify whether \[ P(\text{sum}=7 \text{ and first die}=2) = P(\text{sum}=7) \times P(\text{first die}=2). \] We have \[ P(\text{sum}=7) = \frac{6}{36} = \frac{1}{6} \quad \text{and} \quad P(\text{first die}=2) = \frac{6}{36} = \frac{1}{6}. \] Thus, \[ P(\text{sum}=7) \times P(\text{first die}=2) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}. \] Since the event "sum is 7 and first die is 2" corresponds exactly to the outcome \((2,5)\), \[ P(\text{sum}=7 \text{ and first die}=2) = \frac{1}{36}. \] Because both sides are equal, the events are independent.